g 4 not unimodular, g 3 not unimodular

5.5. MAIN RESULTS 116 with A4,8, A4,10 admit cocalibrated G2-structures by Proposition 5.12 (c) if h¹(g3) ≥ 1 (note thath¹(g₄)−h²(g₄) = 1 for g₄ ∈ {A_4,8, A_4,10} by Table 7.3) and by Corollary 5.14 if h¹(g3) = 0, i.e. g₃∈ {so(3),so(2,1)}. This nishes the proof of Theorem 5.18 (c).

5.5. MAIN RESULTS 117 for alli, j∈N0. Moreover,d(ˆµ) =−tr(F)ˆµ∧e⁴ for allµˆ∈Λ^2,0,0,0andd(˜µ) =−tr(G)˜µ∧e⁷ for all µ˜∈Λ^0,2,0,0.

LetΨ∈Λ⁴(g4⊕g₃)^∗ be the Hodge dual of a cocalibratedG2-structure. DecomposeΨ into

Ψ = Ω +e¹∧ρ

withΩ∈Λ⁴(V₂⊕span(e⁴)⊕g^∗₃),ρ∈Λ³(V₂⊕span(e⁴)⊕g^∗₃). Then

0 =dΨ =dΩ + (tr(F)e¹⁴+ν)∧ρ−e¹∧dρ=e¹∧(tr(F)e⁴∧ρ−dρ) +dΩ +ν∧ρ (5.6) implies Φ := tr(F)e⁴∧ρ−dρ= 0. We look at dierent components of Φ. We have the identities

0 = Φ^2,1,1,0 = tr(F)e⁴∧ρ^2,1,0,0−d(ρ^2,1,0,0)^2,1,1,0 = tr(F)e⁴∧ρ^2,1,0,0−tr(F)ρ^2,1,0,0∧e⁴,

= 2tr(F)e⁴∧ρ^2,1,0,0,

0 = Φ^1,2,0,1 =−d(ρ^1,2,0,0)^1,2,0,1=−tr(G)ρ^1,2,0,0∧e⁷,

0 = Φ^2,0,1,1 = tr(F)e⁴∧ρ^2,0,0,1−d(ρ^2,0,0,1) = 2tr(F)e⁴∧ρ^2,0,0,1, which implyρ^2,1,0,0 =ρ^1,2,0,0 =ρ^2,0,0,1 = 0. Moreover,

0 = Φ^0,2,1,1 = tr(F)e⁴∧ρ^0,2,0,1−d(ρ^0,2,1,0) = tr(F)e⁴∧ρ^0,2,0,1+ tr(G)e⁷∧ρ^0,2,1,0, i.e. ^tr(F_tr(G)⁾e⁴∧ρ^0,2,0,1 =−e⁷∧ρ^0,2,1,0. Thus, ρdecomposes as

ρ=e⁷∧(ω1+ ˆν) +e⁴∧

ω2+tr(F) tr(G)νˆ+λν

+e⁴⁷∧α

with ω₁, ω₂ ∈ Λ^1,1,0,0, νˆ ∈ Λ^0,2,0,0, λ ∈ R, α ∈ Λ^1,0,0,0 ⊕Λ^0,1,0,0. Proposition 2.49 and Lemma 5.15 (b) imply thatω˜1 :=ω1+ ˆνandω˜2 :=ω2+^tr(F_tr(G)⁾νˆ+λν span a two-dimensional subspace in which each non-zero element is of length two. Moreover,

0 = Φ^1,1,1,1= tr(F)e⁴∧ρ^1,1,0,1−d(ρ^1,1,1,0)−d(ρ^1,1,0,1) which shows that

d(e¹∧(ρ^1,1,1,0+ρ^1,1,0,1)) =(ν+ tr(F)e¹⁴)∧(ρ^1,1,1,0+ρ^1,1,0,1)−e¹∧d(ρ^1,1,1,0+ρ^1,1,0,1)

=tr(F)e¹⁴∧ρ^1,1,0,1−e¹∧d(ρ^1,1,1,0)−e¹∧d(ρ^1,1,0,1)

=e¹∧Φ^1,1,1,1 = 0.

Since ρ^1,1,1,0=e⁴∧ω₂ and ρ^1,1,0,1 =e⁷∧ω₁, we getd(ω₁∧e⁷¹+ω₂∧e⁴¹) = 0.

What is left to show is that νˆ 6= 0. Therefore, let Ω˜ be the projection of Ψ onto the subspaceΛ⁴(span(e¹)⊕V₂⊕W₂) (alongP2

i=1Λⁱ(span(e¹)⊕V₂⊕W₂)∧Λ²⁻ⁱspan(e⁴, e⁷)). By Proposition 2.48, l( ˜Ω)≥1, i.e. Ω˜ 6= 0. We may write Ω˜ in terms of the components of ρ and Ωas

Ω =˜ e¹∧ρ^2,1,0,0+e¹∧ρ^1,2,0,0+ Ω^2,2,0,0 = Ω^2,2,0,0

5.5. MAIN RESULTS 118

and get Ω^2,2,0,0 6= 0. Equation (5.6) gives us 0 = (dΩ +ν∧ρ)^2,2,0,1 =d Ω^2,2,0,02,2,0,1

+ν∧ρ^0,2,0,1 =−tr(G)Ω^2,2,0,0∧e⁷+ν∧ρ^0,2,0,1 and so e⁷∧νˆ=ρ^0,2,0,16= 0, i.e. νˆ6= 0.

"⇐":

Assume that there exist two-forms ω₁, ω₂ ∈ V₂∧W₂,0 6= ˆν ∈ Λ²W₂ and λ∈R fullling all the conditions. Then ω˜1 as in the statement fulls 0 6= ˜ω²₁ ∈ Λ²V2∧Λ²W2. Hence, there exists 0 6= ˜λ ∈ R such that ^λ˜₂ω˜₁² = −_tr(G)¹ ν∧νˆ. Set now θ₁ := ¹_˜

λe⁷¹, θ₂ := ¹_˜

λe⁴¹, θ₃ := e⁷⁴ ∈ Λ²span(e¹, e⁴, e⁷). By assumption, ω˜₁, ω˜₂ as in the statement span a two-dimensional space in which each non-zero element has length two. Thus, we may apply Lemma 2.2 and Proposition 2.55 to V₄^∗ := V₂⊕W₂, V₃^∗ := span(e¹, e⁴, e⁷) and get the existence of a two-form ω˜3∈Λ²V₄^∗ such that

Ψ :=

3

X

i=1

˜

ω_i∧θ_i+1 2ω˜₁²

is the Hodge dual of a G2-structure. Using dν = −tr(F)ν∧e⁴, dˆν = −tr(G)ˆν∧e⁷, we compute

dΨ = 1

λ˜d ω˜₁∧e⁷¹+ ˜ω₂∧e⁴¹

+d ω˜₃∧e⁷⁴

− 1

λ˜·tr(G)d(ν∧ν)ˆ

= 1

λ˜d ω1∧e⁷¹+ω2∧e⁴¹ + 1

λ˜d νˆ∧e⁷¹ + 1

λ˜d

tr(F)

tr(G)νˆ∧e⁴¹+λν∧e⁴¹

+ tr(F)

λ˜·tr(G)ν∧νˆ∧e⁴+ 1

λ˜ν∧νˆ∧e⁷

= 0−tr(F)

λ˜ νˆ∧e⁷¹⁴− 1

˜λνˆ∧e⁷∧ν−tr(F)

λ˜ νˆ∧e⁷⁴¹− tr(F)

˜λ·tr(G)νˆ∧e⁴∧ν + tr(F)

λ˜·tr(G)ν∧νˆ∧e⁴+ 1

λ˜ν∧νˆ∧e⁷

= 0.

Remark 5.25. The two-formω₁ ∈V₂∧W₂in Proposition 5.24 has to be of length two since

˜

ω1 =ω1+ˆνis of length two. By Lemma 2.2, there exists a basise², e³ ofV2and a basise⁵, e⁶ ofW2 such thatω1 =e²⁶+e³⁵. If det(G)6= 0, then the conditiond(ω1∧e⁷¹+ω2∧e⁴¹) = 0 implies that ω₂ = (F + tr(F)id)(e²)∧G⁻¹(e⁶) + (F+ tr(F)id)(e³)∧G⁻¹(e⁵).

Let us, nevertheless, start with det(G) = 0.

Lemma 5.26. Let g, g₄, g₃, e¹, e⁴∈g^∗₄, e⁷ ∈g^∗₃, V2, F :V2 →V2, W2 andG:W2→W2

as in Proposition 5.24. Assume further that det(G) = 0, i.e. g₃ =r₂⊕R. Then g admits a cocalibrated G₂-structure if and only if det(F+ tr(F)id) = 0, i.e. g₄ =A⁻

1 2

4,9.

5.5. MAIN RESULTS 119 Proof. "⇒:"

Assume thatgadmits a cocalibratedG₂-structure. By Proposition 5.24 and Remark 5.25, there exists a basis e², e³ of V2 and a basis e⁵, e⁶ of W2 such that ω1 := e²⁶+e³⁵ fulls d(ω1∧e⁷¹)∈d(V2∧W2∧e⁴¹) =V2∧G(W2)∧e⁷⁴¹. Each element inV2∧G(W2)∧e⁷⁴¹ is of length at most one due to det(G) = 0. But

d(ω₁∧e⁷¹) = ((F+ tr(F)id)(e²)∧e⁶+ (F + tr(F)id)(e³)∧e⁵)∧e⁷⁴¹

is of length less than two if and only ifdet(F+ tr(F)id) = 0. Thus,det(F+ tr(F)id) = 0.

"⇐:"

We havedet(F+tr(F)id) = 0 = det(G)andtr(F+tr(F)id) = 3tr(F)6= 0,tr(G)6= 0. Since bothF+tr(F)idandGare linear endomorphisms in two dimensions, this implies that they diagonalisable over the reals with one zero eigenvalue and one non-zero eigenvalue. We may, after rescalinge⁴ and e⁷, assume that the non-zero eigenvalue is equal to one in both cases and sotr(F) = ¹₃ andtr(G) = 1. Sinced(e¹∧α) =−e¹∧(F+ tr(F)id)(α)∧e⁴ for all α∈V₂, there exists a basise², e³ ofV₂ such thatde¹²= 0andde¹³=−e¹³⁴. Moreover, we may choose a basise⁵, e⁶ of W₂ withde⁵= 0 andde⁶ =e⁶⁷. Then the following two-forms full all the conditions in Proposition 5.24:

ω1:=e²⁵−e³⁶+e²⁶, ω2 :=e²⁵−e³⁶−2e³⁵, ω˜1:=e⁵⁶+ω1, ω˜2 := 1

3e⁵⁶+ω2.

If det(G)6= 0 andF and Gare both not multiples of the identity, we get:

Lemma 5.27. Let g, g₄, g₃, e¹, e⁴∈g^∗₄, e⁷ ∈g^∗₃, V₂, F :V₂ →V₂, W₂ andG:W₂→W₂ as in Proposition 5.24. Assume further thatF andGare both not multiples of the identity, i.e. g₄ 6=A¹_4,9 and g₃ 6=r_3,1. Then g admits a cocalibrated G₂-structure.

Proof. SetH :=−(F+ tr(F)id). Then alsoH:V₂ →V₂ is not a multiple of the identity, not trace-free andd(e¹∧α) =e¹∧H(α)∧e⁴ for allα∈V2. By rescaling e⁴ appropriately, we may assume that tr(H) =−3, i.e. tr(F) = 1. Hence, we may choose a basis e², e³ of V₂ such that the transformation matrix ofH with respect to this basis is given by

0 ^det(H)_det(G)

−det(G) −3

! .

Moreover, by rescaling e⁷ appropriately, we may assume that tr(G) = 1. Hence, for all a∈R\{0}, we may choose a basis e⁵, e⁶ of W₂ such that the transformation matrix ofG with respect to this basis is given by

0 −^det(G)_a

a 1

! .

5.5. MAIN RESULTS 120

Set

ω1 :=e²⁵+e³⁶, ω2:=−det(H)

det(G)ae²⁵+3 +a

a e³⁵−a e³⁶, ω˜1 :=e⁵⁶+ω1, ω˜2 :=e⁵⁶−a e²³+ω2. A short computation shows d(ω₁∧e⁷¹+ω₂∧e⁴¹) = 0. Moreover, ω˜²₁ = 2e²⁵³⁶ 6= 0 and

˜

ω1∧ω˜2=Bω˜₁²,ω˜²₂ =Cω˜₁² withB=−_2adet(G)^det(H) and C=a+^det(H_det(G)⁾. Hence, C−B² =a+det(H)

det(G) − det(H)² 4a²det(G)² >0

fora >0large enough and soω˜1,ω˜2 span a two-dimensional space in which each non-zero element has length two by Lemma 2.2. Thus, g admits a cocalibrated G2-structure by Proposition 5.24

Therefore, it remains to consider the cases when at least one of the maps F and G is (a multiple of) the identity:

Lemma 5.28. Let g, g₄, g₃, e¹, e⁴∈g^∗₄, e⁷ ∈g^∗₃, V2, F :V2 →V2, W2 andG:W2→W2

as in Proposition 5.24.

(a) If F is a multiple of the identity, i.e. g₄ = A¹_4,9, then g admits a cocalibrated G₂ -structure if and only if −³₄tr(G)²>det(G) or det(G)>0.

(b) If G is a multiple of the identity, i.e. g₃ = r_3,1, then g admits a cocalibrated G₂ -structure if and only if det(F)>−³₄tr(F)².

Remark 5.29. Note that a real two-by-two matrix with negative determinant is always diagonalisable over the reals. The determinant of G is negative if the condition in Lemma 5.28 (a) is not fullled and the determinant of F is negative if the condition in Lemma 5.28 (b) is not fullled. Hence, it is easily checked that the condition on g₃ in Lemma 5.28 (a) is not fullled exactly when g₃ ∈

r_3,µ µ∈

−¹₃,0

and that the condition on g₄ in Lemma 5.28 (b) is not fullled exactly when g₄ ∈

A^α_4,9

α∈ −1,−¹₃

. Hence, proving Lemma 5.28 nishes the proof of Theorem 5.18.

Proof. (a) By rescaling e⁴ we may assume that tr(F) = 2, i.e. F = id. Hence, Proposi-tion 5.24 and Remark 5.25 tell us that g admits a cocalibratedG2-structure if and only if there exists a basise², e³ ofV₂, a basise⁵, e⁶ ofW₂,λ, α∈R,α6= 0such that each non-zero linear combination of

˜

ω1,α,λ:=αe⁵⁶+e²⁶+e³⁵, ω˜2,α,λ:= 2

tr(G)αe⁵⁶+λe²³+ 3e²∧G⁻¹(e⁶) + 3e³∧G⁻¹(e⁵) is of length two. A short computation shows

˜

ω²_1,α,λ= 2e²³⁵⁶, ω˜_1,α,λ∧ω˜_2,α,λ=

αλ+ 3tr(G) det(G)

e²³⁵⁶,

˜ ω²_2,α,λ=

4 αλ

tr(G) + 18 1 det(G)

e²³⁵⁶

5.5. MAIN RESULTS 121

since for an invertible two-by-two matrix tr G⁻¹

= _det(G)^tr(G). Set X := αλ. Then Lemma 2.2 tells us that each non-zero linear combination of ω˜1,α,λ and ω˜2,α,λ is of length two if and only if the quadratic polynomial

8 X

tr(G)+ 36 1 det(G) −

X+ 3tr(G) det(G)

2

=−X²+ 8

tr(G) −6 tr(G) det(G)

X+ 36 1

det(G) −9 tr(G)² det(G)²

in X with leading negative coecient is positive for some X ∈ R. Note that this expression does not depend on the basis we have chosen. Hence, g admits a cocal-ibrated G2-structure if and only if this quadratic polynomial is positive for some X∈Rand this is true if and only if its discriminant is positive. The discriminant is given by

6 tr(G)

det(G) −8 1 tr(G)

2

−4·

9 tr(G)²

det(G)² −36 1 det(G)

= 16(3tr(G)²+ 4 det(G)) det(G)tr(G)² , and it is positive if and only if

−3

4tr(G)²>det(G) or det(G)>0.

(b) By rescaling e⁷ we may assumetr(G) = 2, i.e. G= id. Then we see similarly as in the proof of part (a) that g admits a cocalibrated G₂-structure if and only if there exists a basise², e³ofV₂, a basise⁵, e⁶ofW₂,λ, α∈R,α6= 0such that each non-zero linear combination of

˜

ω_1,α,λ:=αe⁵⁶+e²⁶+e³⁵,

˜

ω2,α,λ:= tr(F)

2 αe⁵⁶+λe²³+ (F + tr(F)id)(e²)∧e⁶+ (F+ tr(F)id)(e³)∧e⁵ is of length two. If we set X := αλ as before, we nd, analogously to the proof of (a), that the existence of a cocalibrated G₂-structure on g is equivalent to the existence of X ∈R such that−X²−4tr(F)X−tr(F)²+ 4 det(F) is positive. Note therefore that for a two-by-two matrixA∈R^2×2we generally havedet(A+tr(A)I₂) = det(A)+2tr(A)². Now−X²−4tr(F)X−tr(F)²+4 det(F)is positive for someX ∈R exactly when the discriminant of this quadratic polynomial inX, which is given by 12tr(F)²+ 16 det(F), is positive. And this is the case if and only if

det(F)>−3

4tr(F)².

Chapter 6

Half-at structures on Lie algebras

In this chapter, we present the classication results for half-atSU(3)-structures on certain classes of Lie algebras the author obtained together with Fabian Schulte-Hengesbach in the two papers [FS1] and [FS2]. Moreover, we present also some partial results on the classication of six-dimensional Lie algebras admitting other types of half-at structures.

Apart from one result on certain six-dimensional almost Abelian Lie algebras admitting half-at structures of other types, also these results are joint work with Schulte-Hengesbach and already published in [FS1].

More exactly, we nish the classication of the decomposable six-dimensional Lie alge-bras which admit a half-atSU(3)-structure. Therefore, we determine the direct sums of a four-dimensional Lie algebra and of a two-dimensional Lie algebra and the direct sums of a ve-dimensional Lie algebra andRpossessing a half-atSU(3)-structure. These results are all contained in [FS1] and we also present the non-existence results on stable three-forms of certain type and on half-at SU(1,2)- and half-at SL(3,R)-structures on some of the considered decomposable Lie algebras given in [FS1]. Note that the direct sums of two three-dimensional Lie algebras which admit a half-at SU(3)-structure have been deter-mined before by Schulte-Hengesbach in [SH]. The analogous classication has been done by Conti [C1] for the class of six-dimensional nilpotent Lie algebras. We basically use a renement of Conti's and Schulte-Hengesbach's obstructions to prove the non-existence of half-at SU(3)-structures on the mentioned decomposable Lie algebras. Our obstruction has the advantage that it is easy to check using a computer algebra system. In fact, we use Maple, in particular the packages diorms and diorms2, to check the obstruction. Ex-istence is proved in most cases by giving an explicit example of a half-atSU(3)-structure.

We changed parts of the proofs given in [FS1] and use also the relation between half-atSU(3)-structures on six-dimensional Lie algebras gand cocalibrated G₂-structures on g⊕R. We give a direct proof that a six-dimensional almost Abelian Lie algebrag admits a half-at SU(3)-structure if and only if g⊕R admits a cocalibrated G2-structure and

6.1. KNOWN RESULTS AND OBSTRUCTIONS 123 so get a full classication of the six-dimensional almost Abelian Lie algebras admitting a half-at SU(3)-structure by Theorem 5.18. Moreover, we prove that in this case the six-dimensional almost Abelian Lie algebra g admits half-at structures of any type, a result not contained in [FS1] or [FS2]. We also apply the classication of the direct sums of four- and three-dimensional Lie algebras admitting a cocalibrated G₂-structure given in Theorem 5.18 to show that on certain decomposable six-dimensional Lie algebras there cannot exist a half-at SU(3)-structure.

Moreover, we classify the indecomposable solvable six-dimensional Lie algebras with ve-dimensional nilradical which admit a half-atSU(3)-structure and show that all non-solvable six-dimensional Lie algebras possess such a structure. These results are all con-tained in [FS2]. The proofs are completely analogous to the paper [FS2]. We use again our renement of Conti's obstruction but also apply some obstruction obtained by the relation between half-atSU(3)-structures on a six-dimensional Lie algebragand cocalibrated G2 -structures on g⊕R. Existence is again proved by giving concrete examples. Note that by a result of Mubarakzyanov [Mu6d], a six-dimensional solvable indecomposable Lie algebra is nilpotent or the nilradical has dimension ve or four. Hence, only the question which in-decomposable solvable six-dimensional Lie algebras with four-dimensional nilradical admit a half-at SU(3)-structure remains open. We emphasise that we rened the classication of indecomposable ve-dimensional Lie algebras given in [Mu5d] and also the classication of six-dimensional Lie algebras with ve-dimensional non-Abelian nilradical in [Mu6d] in order to obtain the mentioned classication results. This renement is interesting in its own. We give an application of this renement to the classication of six-dimensional (2,3)-trivial Lie algebras which is also contained in [FS2].

We start in Section 6.1 by giving a brief history of the results known before and also of the obstructions used by Conti in [C1] and by Schulte-Hengesbach in [SH] to obtain their results. Section 6.2 presents our renement of these obstructions and also the above-mentioned obstruction obtained by the relation between half-at SU(3)-structures on a six-dimensional Lie algebra gand cocalibrated G₂-structures ong⊕R. In Section 6.3, we prove the classication results on six-dimensional Lie algebras admitting half-at SU(3) -structures. Finally, Section 6.4 gives the result on the classication of six-dimensional (2,3)-trivial Lie algebras. Moreover, also the non-existence results on stable forms of certain kind and on half-atSU(1,2)- orSL(3,R)-structures in the decomposable case are presented in this section.

6.1 Known results and obstructions

The rst steps towards a classication of the Lie algebras which admit half-at SU(3) -structures have been done in [ChiSw], [ChiFi], [CT]. In these papers, a classication of the

6.1. KNOWN RESULTS AND OBSTRUCTIONS 124 nilpotent six-dimensional Lie algebras admitting special kinds of half-atSU(3)-structures has been given. The next step has been the following classication of the nilpotent Lie algebras admitting an arbitrary half-atSU(3)-structures [C1] by Conti. For the names of the appearing Lie algebras, we refer the reader to the appendix.

Theorem 6.1 (Conti). Letg be a six-dimensional nilpotent Lie algebra. Then g admits a half-at SU(3)-structure if and only if

(i) g is decomposable and g∈

R⁶,h₃⊕R³,h₃⊕h₃, A_5,i⊕R

i= 1,2,4,5,6} or (ii) g is indecomposable and g=n_6,j for j /∈ {1,2,8,9,10,19,21,22}.

To prove Theorem 6.1, he introduces the concept of a coherent splitting on an arbitrary six-dimensional Lie algebra g, which is a splitting g^∗ = V₂ ⊕V₄ into a two-dimensional subspaceV2 and a four-dimensional subspaceV4 with d(V2) = Λ²V2 and d(V4) = Λ²V2⊕ V₂∧V₄. This splitting can be used to dene a double complex(Λ^p,qg^∗, δ₁, δ₂). Conti shows that the triviality of the cohomology classes H^0,3 andH^0,4 implies that ong there cannot exist a half-at SU(3)-structure. He applies this obstruction then to eight nilpotent Lie algebras to exclude half-at SU(3)-structures on them. The existence is proved by giving a concrete example of a half-atSU(3)-structure in each case. There are two nilpotent Lie algebras, namelyn_6,21 andn_6,22, which do not admit a coherent splitting and existence of half-at SU(3)-structures on them is excluded by rening the methods. Basically, Conti uses Equation (2.16) and Equation (2.17). More exactly, he shows that the existence of a half-at SU(3)-structure(ω, ρ) ∈Λ²g^∗×Λ³g^∗ implies in both cases that the basis vector e¹ in the basis given in Table 7.6 fulls J_ρ^∗e¹ ∈span(e¹, e², e³). For that purpose he uses Equation (2.16), which states that

e¹∧(vyρ)∧ρ=J_ρ^∗e¹(v)φ(ρ)

for all v ∈ g, and shows that e¹∧(wyψ)∧ψ = 0 for all closed three-forms ψ and all w∈span(e₄, e₅, e₆). Afterwards, he computes that then the identitye¹∧J_ρ^∗e¹∧σ = 0for all closed four-forms σ ∈Λ⁴g^∗, and so also for ω², is true. But this is a contradiction to Equation (2.17), which states that

β∧J_ρ^∗β∧ω² = 1

3g(β, β)ω³ 6= 0 for all β∈g^∗\{0}.

Our obstruction given in Proposition 6.5 resembles this argumentation. It gives a direct obstruction for which one has to compute all closed three-forms and all closed four-forms on g and can then easily check the obstruction. The obtained obstruction is built up in such a way that all computations can be done using a computer algebra system like Maple.

6.1. KNOWN RESULTS AND OBSTRUCTIONS 125 The next class of Lie algebras considered was the direct sums of two three-dimensional Lie algebras by Schulte-Hengesbach in [SH]. He gets an obstruction without introducing the double complex above by looking at the decisive steps in Conti's proof. He applies this obstruction to all but two cases of direct sums g =g₁ ⊕g₂ of two three-dimensional Lie algebras g₁,g₂ which do not admit a half-at SU(3)-structure. Existence is again proved by giving concrete examples of half-at SU(3)-structures. The missing two cases r₂⊕R⁴ and h₃ ⊕r₂ ⊕R are treated separately. The rst case is excluded by showing that all closed three-formsρ∈Λ³ r₂⊕R⁴∗

fullλ(ρ)≥0and sor₂⊕R⁴ cannot admit a half-at SU(3)-structure by Proposition 3.38. The second case uses directly our main obstruction given below in Proposition 6.5 without stating it concretely.

The result obtained by Schulte-Hengesbach is given in Theorem 6.2. We rephrase it to make connection to the existence of cocalibratedG₂-structures on direct sumsg₁⊕g₂⊕R withdim(gi) = 3 for i= 1,2.

Theorem 6.2 (Schulte-Hengesbach). A direct sum g₁⊕g₂ of two three-dimensional Lie algebrasg₁,g₂ admits a half-atSU(3)-structure if and only if both g₁ andg₂ are unimod-ular or exactly one of the Lie algebras is unimodunimod-ular, say g_i1 for some i1 ∈ {1,2}, and h²(gi1)≤h²(gi2), where i2∈ {1,2} is such that{i₁, i2}={1,2}.

Remark 6.3. Theorem 6.2 and Theorem 5.18 imply that a direct sumg₁⊕g₂ of two three-dimensional Lie algebras g₁,g₂ admits a half-atSU(3)-structure if and only ifg₁⊕g₂⊕R admits a cocalibrated G2-structure. By Proposition 3.37, the existence of a half-at SU(3) -structure on a six-dimensional Lie algebra g is equivalent to the existence of a cocalibrated G2-structure on g⊕Rsuch that gis orthogonal to R. Hence, the non-existence of half-at SU(3)-structure in all cases in Theorem 6.2 follows independently also by our Theorem 5.18. However, the existence of half-at SU(3)-structures on the Lie algebras in Theorem 6.2 does not follow by Theorem 5.18 since we cannot ensure the existence of a cocalibrated G₂-structure such that g₁ ⊕g₂ is orthogonal to R. In fact, in Remark 6.10 we present an example of a direct sum g = g₄⊕g₂ of a four-dimensional Lie algebra g₄ and a two-dimensional Lie algebra g₂ which does not admit a half-at SU(3)-structure but for which g₄⊕g₂⊕Radmits a cocalibrated G2-structure.

Schulte-Hengesbach also classied the direct sums g₁⊕g₂ of two three-dimensional Lie algebras admitting a half-atSU(3)-structure such that the decomposition is orthogonal.

Moreover, he also classied the direct sumsg₁⊕g₂ admitting a half-atSL(3,R)-structure for which the decomposition is orthogonal and each summand is denite and did the same under the condition that the summands are the±1-eigenspaces of the para-complex structure. We do not give these classications here. Instead, we mention his results that certain direct sums do not admit any half-at SU(1,2)-structure. He achieved this result by showing λ(ρ) ≥ 0 for all closed three-forms using Maple for the calculations.

6.2. NEW OBSTRUCTIONS 126

Im Dokument Geometric structures on Lie algebras and the Hitchin flow (Seite 134-144)